What is the change in heat energy when 75 Kg of water at 9.8°C is raised to 22.4°C?

4,000 Kilojoules (2 sig. figs. based on 2 sig. figs. in 75Kg water).    

Explanation:

Energy change equation => q = m·c·ΔT

q = total amount of heat flow

m = mass in grams = 75,000 grams

c = specific heat of water = 4.184 joules/gram·°C

ΔT = Temperature change in °C = 22.4°C - 9.8°C = 12.6°C

q =(75,000g)(4.184j/g·°C)(12.6°C) = 3,953,880 joules = 3,954 Kilojoules ≅ 4,000 Kilojoules (2 sig. figs. based on 2 sig. figs. in 75Kg water).

i think   the answer is c hope i'm right

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2